# Thread: MATH2111 Higher Several Variable Calculus

1. ## Re: Multivariable Calculus

I'm so shit at this topic.

$Let F(u,v)=\int_1^u{g(v+2t)dt} where g has a continuous derivative.$

$The answers found \frac{\partial F}{\partial u}=g(v+2u) using FTC1 which I perfectly get$

But then they find \\ \begin{align*}\frac{\partial F}{\partial v}&=\int_1^u{g^\prime (v+2t) dt} \quad \text{which appears ok at first sight; it's just Leibniz's rule}\\ =\left[\frac{1}{2}g(v+2t)\right]_1^u\end{align*}
____________

$So like, this only partially makes sense. The g^\prime was forced out by taking a derivative with respect to v. So why is it ok that when integrating with respect to t, the prime can get dropped off again?$

Edit: Why is LaTeX broken on this site again?

Ok basically the main question is

They had d/dv g(v+2t) = g'(v+2t)

So why was it safe to integrate with respect to t to get 1/2 g(v+2t) again? I'm missing something elementary

2. ## Re: Multivariable Calculus

Originally Posted by leehuan
I'm so shit at this topic.

$Let F(u,v)=\int_1^u{g(v+2t)dt} where g has a continuous derivative.$

$The answers found \frac{\partial F}{\partial u}=g(v+2u) using FTC1 which I perfectly get$

But then they find \\ \begin{align*}\frac{\partial F}{\partial v}&=\int_1^u{g^\prime (v+2t) dt} \quad \text{which appears ok at first sight; it's just Leibniz's rule}\\ =\left[\frac{1}{2}g(v+2t)\right]_1^u\end{align*}
____________

$So like, this only partially makes sense. The g^\prime was forced out by taking a derivative with respect to v. So why is it ok that when integrating with respect to t, the prime can get dropped off again?$

Edit: Why is LaTeX broken on this site again?

Ok basically the main question is

They had d/dv g(v+2t) = g'(v+2t)

So why was it safe to integrate with respect to t to get 1/2 g(v+2t) again? I'm missing something elementary
They're integrating the derivative of a function, so they used FTC. (The 1/2 coming out due to reverse chain rule basically.)

Note that (∂/∂v)(g(v+2t)) = g'(v+2t), where g' just means the derivative of the function g.

E.g. if g(y) was cos(y), we'd have ∂/∂v (g(v+2t)) = ∂/∂v (cos(v+2t)) = -sin(v+2t) (i.e. g' evaluated at v+2t, which is what g'(v+2t) means).

Then integrating -sin(v+2t), we'd get back to (1/2)cos(v+2t).

3. ## Re: Multivariable Calculus

Originally Posted by InteGrand
They're integrating the derivative of a function, so they used FTC. (The 1/2 coming out due to reverse chain rule basically.)

Note that (∂/∂v)(g(v+2t)) = g'(v+2t), where g' just means the derivative of the function g.

E.g. if g(y) was cos(y), we'd have ∂/∂v (g(v+2t)) = ∂/∂v (cos(v+2t)) = -sin(v+2t) (i.e. g' evaluated at v+2t, which is what g'(v+2t) means).

Then integrating -sin(v+2t), we'd get back to (1/2)cos(v+2t).
Oops thanks.

It makes sense with an example; I think I just got scared of too many variables here

4. ## Re: Multivariable Calculus

Originally Posted by leehuan
I'm so shit at this topic.

$Let F(u,v)=\int_1^u{g(v+2t)dt} where g has a continuous derivative.$

$The answers found \frac{\partial F}{\partial u}=g(v+2u) using FTC1 which I perfectly get$

But then they find \\ \begin{align*}\frac{\partial F}{\partial v}&=\int_1^u{g^\prime (v+2t) dt} \quad \text{which appears ok at first sight; it's just Leibniz's rule}\\ =\left[\frac{1}{2}g(v+2t)\right]_1^u\end{align*}
____________

$So like, this only partially makes sense. The g^\prime was forced out by taking a derivative with respect to v. So why is it ok that when integrating with respect to t, the prime can get dropped off again?$
In the integrand we are applying g (a function of a single variable) to the expression v+2t.

Partially differentiating g(v+2t) with respect to v and partially differentiating with respect to t just differ by a factor of 2, as computed using the chain rule. (Write h(v,t)=v+2t as a function from R^2 to R and carefully apply the chain rule to the composition (g o h) if you are still confused after this post.)

So to summarise, Leibniz lets us differentiate the integral w.r.t. v by differentiating the integrand w.r.t. v. The new integrand can be replaced by the t-partial derivative of g(v+2t) up to the constant factor of 2 which we account for by division. Then we are just integrating the t-derivative of something w.r.t t which the FTC takes care of.

5. ## Re: Multivariable Calculus

Another exercise without much required knowledge.

Suppose we have a smooth function $X:\mathbb{R}^n\rightarrow \mathbb{R}^n$.

Suppose further, that given any starting point $x\in\mathbb{R}^n$, we can find a smooth curve $c(t)$ with $c(0)=x$ and $c'(t)=X(c(t))$.

(Intuitively, the vector field X is the infinitesimal generator of the curve c.)

Under some simple assumptions, the function $p_t(x): \mathbb{R}^n \rightarrow \mathbb{R}^n$ that maps a pair $x$ to $c(t)$ where c is the curve starting at x, is a smooth function with smooth inverse for any fixed t.

So we can consider how sets in R^n "flow". Of course, in general this will distort volume and other geometric quantities.

One way of quantifying volume distortion locally about a point p is as follows:

$E_X(p):= \lim_{\epsilon\rightarrow 0}\frac{\frac{d}{dt}(|p_t(B(p,\epsilon))|)|_{t=0}} {|B(p, \epsilon)|}$

where |S| denotes the volume of a subset S in R^n, or equivalently, the integral of the constant function 1 over this set, which we assume is a well defined Riemann integral for the sets involved in this question and B(p,r) denotes the ball about p of radius r.

(So E_X measures the limiting rate of change of volumes of small balls about p relative to the size of these balls.)

1. Compute an expression for E_X(p) in terms of the components of X and it's derivatives.

Followups to come if anyone answers this question.

6. ## Re: Multivariable Calculus

Originally Posted by leehuan
So like, this question felt never ending and I aborted.

Part b) is just the chain rule:

$b) Show that \frac{\partial f}{\partial x}=\cos{\theta}\frac{\partial f}{\partial r}-\frac{\sin\theta}{r} \frac{\partial f}{\partial \theta} and similar for y$

Is there any shortcut to save me from computing several product and quotient rules in this one

$c) Show that \\ \frac{\partial^2 f}{\partial x^2}=\frac{\sin^2{\theta}}{r}\frac{\partial f}{\partial r}+\frac{2\sin{\theta}\cos{\theta}}{r^2} \frac{\partial f}{\partial \theta}+\cos^2{\theta} \frac{\partial^2 f}{\partial r^2} - \frac{2\cos{\theta} \sin\theta}{r}\frac{\partial^2 f}{\partial r \partial \theta} + \frac{\sin^2{\theta}}{r^2} \frac{\partial^2 f}{\partial \theta^2}$
Aha thanks InteGrand, I gave this question much more thought today and finally got the proof out (one and a half months late :P )

7. ## Re: Multivariable Calculus

Three months later and I still don't know how to use total differential approximation. Correct answer = 0.21% (possibly approximated)

$A triangle has two sides of length a and b with an included angle measuring \frac{\pi}{3} radians. Given that a increases by 5\%, b decreases by 6\%, and the included angle increases by 2\%, estimate the percentage increase of area of the triangle.$

Deductions so far:\\ \begin{align*}\frac{\partial A}{\partial a}&=\frac{1}{2}b\sin C\\ \frac{\partial A}{\partial b}&=\frac{1}{2}a\sin C\\ \frac{\partial A}{\partial C}&=\frac{1}{2}ab\cos C\end{align*}

8. ## Re: Multivariable Calculus

Originally Posted by leehuan
Three months later and I still don't know how to use total differential approximation. Correct answer = 0.21% (possibly approximated)

$A triangle has two sides of length a and b with an included angle measuring \frac{\pi}{3} radians. Given that a increases by 5\%, b decreases by 6\%, and the included angle increases by 2\%, estimate the percentage increase of area of the triangle.$

Deductions so far:\\ \begin{align*}\frac{\partial A}{\partial a}&=\frac{1}{2}b\sin C\\ \frac{\partial A}{\partial b}&=\frac{1}{2}a\sin C\\ \frac{\partial A}{\partial C}&=\frac{1}{2}ab\cos C\end{align*}
Just find the equation of the tangent hyper-plane at the point ½absinC and plug in the given values...

Here, the tangent hyperplane is given by:

$\noindent w(a,b,C) = \frac{b \sqrt{3}}{4}(a_\Delta - a) + \frac{a \sqrt{3}}{4}(b_\Delta - b) + \frac{ab}{4}(C_\Delta - \frac{\pi}{3}) + \frac{ab \sqrt{3}}{4} \\ The differential approximation will be given by plugging in the specified values of a_\Delta, b_\Delta, C_\Delta$

9. ## Re: Multivariable Calculus

Just find the equation of the tangent hyper-plane at the point ½absinC and plug in the given values...

Here, the tangent hyperplane is given by:

$\noindent w(a,b,C) = \frac{b \sqrt{3}}{4}(a_\Delta - a) + \frac{a \sqrt{3}}{4}(b_\Delta - b) + \frac{ab}{4}(C_\Delta - \frac{\pi}{3}) + \frac{ab \sqrt{3}}{4} \\ The differential approximation will be given by plugging in the specified values of a_\Delta, b_\Delta, C_\Delta$
I'm asking for a friend. Can we just break it back down to appropriate use of the formula thanks.

10. ## Re: Multivariable Calculus

Originally Posted by leehuan
I'm asking for a friend. Can we just break it back down to appropriate use of the formula thanks.
it's the higher dimensional analogue of construction of a tangent at a point. fairly intuitive to understand if you ask me, and also I have no knowledge of what constitutes "appropriate use"

11. ## Re: Multivariable Calculus

I don't want my friends to just see the word "hyperplane" and run off. (Nor do I have enough of a grasp to translate it myself, even if it seems to make perfect sense to me)

$By formula I meant using this, but it didn't work when I tried\\ \Delta A = \frac{\partial A}{\partial a}\Delta a+\frac{\partial A}{\partial b}\Delta b+\frac{\partial A}{\partial C}\Delta C$

(think I meant basic. No idea where I got 'appropriate' from.)

12. ## Re: Multivariable Calculus

$\noindent Note A = \frac{1}{2}ab \sin C. Percentage change is given by \frac{\Delta A}{A}. Note \Delta A \approx \frac{1}{2}b \sin C \Delta a + \frac{1}{2}a \sin C \Delta b + \frac{1}{2}ab \cos C \Delta C (total differential approximation). So dividing through by A = \frac{1}{2} a b \sin C, we get an answer of \frac{\Delta A}{A} \approx \frac{\Delta a}{a} + \frac{\Delta b}{b} + \cot C \Delta C = 0.05 -0.06 + \cot \frac{\pi}{3}\cdot \left(0.02 \times \frac{\pi}{3}\right) = 0.0020920 \ldots \approx 0.21\%.$

$\noindent The hyperplane thing is essentially what the total differential approximation is -- we're approximating the change in the function A by moving along its tangent hyperplane by the amounts the variables changed by. It's a higher dimensional analog of the tangent-line approximation from single-variable calculus f\left(x+\Delta x \Right) \approx f(x) + f^\prime (x) \Delta x (or written another way, \Delta f \approx f^\prime (x)\Delta x). The reason it's a tangent hyperplane' and not just a tangent line or tangent plane is that there are three independent variables in this question, so the graph of the function lives in four dimensions, with the tangent hyperplane being a three-dimensional object.$

13. ## Several Variable Calculus Marathon & Questions

For some reason I lost the point of intersection (2,0):

$\text{Find the }\textit{two}\text{ points of intersection of the curves}\\ \textbf{r}(t)=(t^2-t,t^2+t)\\ \textbf{r}(t)=(t+t^2,t-t^2)$

I equated the relevant r1 and r2 components but only got t=0

14. ## Re: Several Variable Calculus

Originally Posted by leehuan
For some reason I lost the point of intersection (2,0):

$\text{Find the }\textit{two}\text{ points of intersection of the curves}\\ \textbf{r}(t)=(t^2-t,t^2+t)\\ \textbf{r}(t)=(t+t^2,t-t^2)$

I equated the relevant r1 and r2 components but only got t=0
Solve the system of equations

t^2 - t = s + s^2 (1)
t^2 + t = s - s^2 (2).

15. ## Re: Several Variable Calculus

Originally Posted by InteGrand
Solve the system of equations

t^2 - t = s + s^2 (1)
t^2 + t = s - s^2 (2).
Oh. I'll work on that right now but why was it necessary to introduce a new variable?

16. ## Re: Several Variable Calculus

Originally Posted by leehuan
Oh. I'll work on that right now but why was it necessary to introduce a new variable?
You have two parametric curves. Their points of intersection don't necessarily have the same parameter as a point on the first curve as they do as a point on the second curve.

17. ## Re: Several Variable Calculus

Eg, consider the lines (x,y)=(t,0), (x,y)=(2t,0).

These lines coincide exactly, so every point on the x-axis is a point of intersection.

Yet the only place where (t,0)=(2t,0) is at the origin.

18. ## Re: Several Variable Calculus

Excellent. Makes sense.
________________

Find the angle between the two curves at the points of intersection.

I'm having a dumb moment now. Which vectors are we taking the dot product of?

19. ## Re: Several Variable Calculus

Originally Posted by leehuan
Excellent. Makes sense.
________________

Find the angle between the two curves at the points of intersection.

I'm having a dumb moment now. Which vectors are we taking the dot product of?
Find the s and t values at the points of intersection and plug them into the derivatives of the parametric curves. This will give us the "direction vectors" of the curves at the points of intersection. Find the angle between these vectors.

20. ## Re: Several Variable Calculus

Strange... That's what I did so maybe there's an error in my computation.

$\text{The values for }(s,t)\text{ were }(1,-1)\text{ and }(0,0)\text{ and the latter easily worked}$

$\\ \textbf{r}_1^\prime (t) = (2t-1,2t+1)\\ \textbf{r}_2^\prime (s) = (1+2s, 1-2s)$

$\text{So }\\\textbf{r}_1^\prime (-1)=(-5,-1)\\ \textbf{r}_2^\prime(1)=(5,-1)$

\begin{align*}\textbf{r}_1^\prime\cdot \textbf{r}_2^\prime&=|\textbf{r}_1^\prime||\textbf{r}_2^\prime|\cos \theta\\ -25 + 1 &= \sqrt{26}\sqrt{26}\cos \theta\end{align*}

EDIT: Ouch. I know what I did now.

21. ## Re: Several Variable Calculus

$\noindent So e.g. the first curve is \mathbf{r}_{1}(t) = \left(t^{2} -t, t^{2} + t\right), so the derivative is \mathbf{r}_{1}^{\prime}(t) = \left(2t-1, 2t+1\right). This direction of this essentially tells us the direction of the curve at t (it's basically the velocity vector of a particle whose position at time t is \mathbf{r}_{1}(t)). So for example at the intersection point (0,0) (which is where t=0 for \mathbf{r}_{1}(t)), the velocity is (-1,1) (subbing t=0 into the `velocity'' function found above). Find the velocity vector also for the other curve at the origin (done in a similar manner), and the find the angle between that vector and (-1,1).$

22. ## Re: Several Variable Calculus

Originally Posted by leehuan
Strange... That's what I did so maybe there's an error in my computation.

$\text{The values for }(s,t)\text{ were }(1,-1)\text{ and }(0,0)\text{ and the latter easily worked}$

$\\ \textbf{r}_1^\prime (t) = (2t-1,2t+1)\\ \textbf{r}_2^\prime (s) = (1+2s, 1-2s)$

$\text{So }\\\textbf{r}_1^\prime (-1)=(-5,-1)\\ \textbf{r}_2^\prime(1)=(5,-1)$

\begin{align*}\textbf{r}_1^\prime\cdot \textbf{r}_2^\prime&=|\textbf{r}_1^\prime||\textbf{r}_2^\prime|\cos \theta\\ -25 + 1 &= \sqrt{26}\sqrt{26}\cos \theta\end{align*}

You appear to have miscalculated the velocity vectors when subbing in the values of t and/or s (check the first components, noting you're subbing in s = 1 (not 2) and t = -1 (not -2)).

23. ## Re: Several Variable Calculus

$\text{Proven earlier: For a particle with velocity }\textbf{v}(t)\\ \text{If }\textbf{v}(t)\cdot \textbf{v}^\prime(t)=0\text{ then the speed }v=\sqrt{v_1^2+\dots+v_n^2}\text{ is constant}$

$\\ \text{A particle of mass }m\text{ with position vector }\textbf{r}(t)\text{ at time }t\text{ is acted on by a total force}\\ \textbf{F}(t)=\lambda \textbf{r}(t)\times \textbf{v}(t)\\ \text{where }\lambda\text{ is a constant and }\textbf{v}(t)\text{ is the velocity of the particle.}$

$\text{Show that the speed }v\text{ of the particle is constant.}$

Can be assumed: F=ma

24. ## Re: Several Variable Calculus

Originally Posted by leehuan
$\text{Proven earlier: For a particle with velocity }\textbf{v}(t)\\ \text{If }\textbf{v}(t)\cdot \textbf{v}^\prime(t)=0\text{ then the speed }v=\sqrt{v_1^2+\dots+v_n^2}\text{ is constant}$

$\\ \text{A particle of mass }m\text{ with position vector }\textbf{r}(t)\text{ at time }t\text{ is acted on by a total force}\\ \textbf{F}(t)=\lambda \textbf{r}(t)\times \textbf{v}(t)\\ \text{where }\lambda\text{ is a constant and }\textbf{v}(t)\text{ is the velocity of the particle.}$

$\text{Show that the speed }v\text{ of the particle is constant.}$

Can be assumed: F=ma
$\noindent Since there's a cross product, we are in \mathbb{R}^{3} here. We have \mathbf{F} (t) = m\mathbf{v}'(t) = \lambda \mathbf{r}(t)\times \mathbf{v}(t). Dotting both sides with \mathbf{v}(t) yields m\mathbf{v}'(t) \cdot \mathbf{v}(t) = 0, which implies by the earlier proven result that the speed is constant.$

25. ## Re: Several Variable Calculus

$\noindent (Note that \left(\mathbf{r}\times \mathbf{v}\right)\cdot \mathbf{v} = 0, recalling that in general a scalar triple product like this is always zero if two of the vectors are the same.)$

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